Yes they should add to 1 because those are the only outcomes. b) Are the outcomes in the sample space or part (a) equally likely? No. The seeds are more likely to germinate. c) Use relative frequencies to estimate the probability that a seed will germinate.
Lauren Sapsed - Essay Question Holmes and Rhae (1967) conducted a study of which they asked 394 people to assess the personal impact or stress value of 43 life events. Death of a partner was rated as having the most impact and was given a value of 100. Then other events were scaled in relation to death of a partner. Adding up the values for events experienced in the previous year gives a life change score for an individual. Others examples from the study social readjustment rating scale (SRRS), divorce 73, redundancy 47, beginning school and a holiday being 13.
RES/342--- This produced a 28/30 on the final 1) What are the critical z-values for a two-tailed hypothesis test if the significant level = 0.01? C. ± 2.58 2) In classical hypothesis testing, the test statistic is to the critical value what the __________. A. ‘p-value’ is to alpha 3) For a hypothesis test of a single population mean at 95% confidence level, a calculated Z score of 1.7 supports the conclusion that A. the population mean is greater than the hypothesized value 4) If the paired differences are normal in a test of mean differences, then the distribution used for testing is the C. student distribution 5) One hundred women were polled and 60 reported successfully communicating an automobile problem to an auto repairman. A sample of 150 men had 95 reporting the same success.
Find the mean and the standard deviation of the sampling distribution of the sample mean of bar x. Mean = 20 (given in the question) Standard deviation = 4/√64 Standard deviation = 4/ 8 Standard deviation = 0.5 c. Calculate the probability that we will obtain a sample mean greater than 21; that is, calculate P(x > 21). Hint: Find the z value corresponding to 21 by using μ and σ because we wish to calculate a probability about x. Then sketch the sampling distribution and the probability Z = (21-20)/ (4/√64) Z = 1/0.5 Z = 2 P (z>2) P = 1-0.9772 P = 0.0228 d. Calculate the probability that we will obtain a sample mean less than 19.385 ; that is calculate P(x.< 19.385). Z = (19.385 - 20)/ (4/√64) Z = -0.615 / 0.5 Z = -1.23 P = 0.1093 7.30 On February 8, 2002, the Gallup Organization released the results of a poll concerning American attitudes toward the 19th Winter Olympic Games in Salt Lake City, Utah.
Estimations for mortality rates are based on five- and ten-year interval ranging from birth to age 70. Since life expectancy ( LE) represents events very early or late in life, S70, represents age of death, and main measurement for mortality rates in the 40's, 50's and 60's. Researchers designed S70 for each sex-race group in every county, however Compressed Mortality Files (CMF) requires a minimum of 2000 total sub population in each area. Due to rules and regulations for privacy the Census defines Public Use Microdata Areas (PUMAs) are intended to capture 100,000+ total population areas. The low density areas of the study were clumped together while high-density counties were sub-divided.
If the value of x is postnatal age of 30 hours, what is the value for Yˆ or SBP for neonates ≤1,000 grams? Show your calculations. 208 EXERCISE 27 • Simple Linear Regression 6. In the legend beneath Figure 2, the authors give an equation indicating that systolic blood pressure is SBP = 50.3 + 0.12x. If the value of x is postnatal age of 30 hours, what is the value for Yˆ or SBP for neonates 1,001–1,500 grams?
This meant that however many chickens or pigs there were, the other would have the number that when combined makes 70 (ex. If there were 2 pigs there would be 68 chickens). Like I said, I already knew that whatever number I put in for chickens would have to make 70 when added to the number of pigs and vise-versa, so my first thought was to try numbers like 10 and 60 or 20 and 50. Below you'll see my table. The Farm Yard Problem Table Chickens | Pigs | Legs | Heads | 0 | 70 | 280 | 70 | 10 | 60 | 260 | 70 | 20 | 50 | 240 | 70 | 30 | 40 | 220 | 70 | 40 | 30 | 200 | 70 | 50 | 20 | 180 | 70 | 60 | 10 | 160 | 70 | 70 | 0 | 140 | 70 | As you can see in my table above, there are 40 chickens and 30 pigs to total the 200 legs and 70 heads.
The same holds true for Han. If Coa chooses to charge limit prices, Han can earn $3 billion rather than $2 billion by charging limit prices. If Coa chooses to charge monopoly prices, Han can earn $4 billion rather than $3 billion by charging limit prices. Irrespective of the pricing strategy chosen by Coa, Han is better off charging limit prices. Limit pricing is a dominant strategy for Han.
According to Exhibit 7a, Samsung’s prices per chip are on average higher by $0.72 than those of its competitors, or by 14.5%. Remember that price is not the same as WTP, it’s usually considerably lower especially in competitive industries. But Samsung’s higher prices do suggest that customers have a higher WTP for Samsung’s chips. What explains the price premium? P.2 mentions that OEMS would an “upwards of 1%” premium for superior reliability, which Samsung surely offers, but that’s only 1% (5-6 cents), not 14.5%.
The first class in a relative frequency table is 50–59 and the corresponding relative frequency is 0.2. What does the 0.2 value indicate? Answer: 0.2 is equal to 1/5 or 20%, 0.2 indicates 20% of the data values are in this particular interval. 3. When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6.